A convolution integral of airy functions

Question

I wonder whether the following integral of Airy functions can be computed? \begin{equation} F(x,y):=\int_{-\infty}^\infty \int_{-\infty}^\infty Ai(x-u)Ai(y-v) e^{ituv}du dv,\quad t \in \mathbb R. \end{equation} It is the convolution of $Ai(x)Ai(y)$ and $e^{itxy}$. (As $Ai(x)$ satisfies the ODE $z^{\prime\prime}-zx=0$, $F(x,y)$ might be the solution of a good PDE; identifying the PDE and its general solutions may help with solving the integral explicitly.)

[An ideal answer would be in a form like $Ai(x+y)Bi(x^2+y^{\frac{1}{2}})$, that is, expressed with usual special functions evaluated at algebraic expressions of $x, y$. ]

Answer 1

I do not know how to solve this problem, but I hope this (comment) would be helpful in someway.

One can use the Airy function's integral expression $$ Ai(x) = \frac{1}{2\pi i} \int_{\mathcal{C}} e^{\frac{T^3}{3} - T x} dT $$ where $\mathcal{C}$ is a contour running from $\infty e^{- i \pi /3}$ to $\infty e^{ i \pi /3}$, or any other contour which can be deform from this one such that $Re(T^3) \to -\infty$ along the contour. After plug-in to the original integral, one can integrate out $u,v$, which would result in an integral of the form $$ C \int_{\mathcal{C}}\int_{\mathcal{C}} e^{\frac{T_1^3 + T_2^3}{3} - T_1 x - T_2 y + c \frac{T_1 T_2}{t}} dT_1 dT_2 $$ where $c$ is some constant. One can then let $T_1 = S - D$ $T_2 = S + D$, where we have deform the contour for $T_i$ to be $\mathcal{C}': a + i \mathbb{R}$ for some $a>0$, and the contour for $S$ is $\mathcal{S} = \mathcal{C}'$, for $D$ is just $i \mathbb{R}$. After making such substitution, the exponent is quadratic in $D$: $(2 S + c/t) D^2 + \cdots$. Performing the Gaussian integral in $D$ again (assuming convergence), one end up with an integral in $S$ of the following form $$ C \int_{\mathcal{C}} e^{f(S)} \frac{1}{\sqrt{2 S + c/t}} d S$$ where $f(S)$ is a cubic polynomial. I am not sure if this can be further simplified.